Question

A manufacturer makes two types of toys $A$ and $B$. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

Types of Toys	Machines	Machines	Machines
	I	II	III
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

Each machine is available for a maximum of $6$ hours per day. If the profit on each toy of type $A$ is $Rs.7.50$ and that on each toy of type $B$ is $Rs.5$, show that $15$ toys of type $A$ and $30$ of type $B$ should be manufactured in a day to get maximum profit.

Answer 1

Let's assume that number of toys of type A be X and number of toys of type B be Y.

Since, toy A need 12 minutes and toy B need 6 minutes of machine $I$ time. Also machine $I$ is available for maximum 6 hours ( 360 minutes ).

So, $12X+6Y\le 360$

$\Rightarrow 2X+Y\le 60...\left(1\right)$

Since, toy A need 18 minutes and toy B need 0 minutes of machine $II$ time. Also machine $I$ is available for maximum 6 hours ( 360 minutes ).

So, $18X+0Y\le 360$

$\Rightarrow X\le 20...\left(2\right)$

Since, toy A need 6 minutes and toy B need 9 minutes of machine $III$ time. Also machine $III$ is available for maximum 6 hours ( 360 minutes ).

So, $6X+9Y\le 360$

$\Rightarrow 2X+3Y\le 120...\left(3\right)$

Since, count of toys can't be negative.

$\therefore X\ge 0andY\ge 0...\left(4\right)$

Now, profit on toy A is $7.50Rs$ and profit on toy B is $5Rs$

So, total profit $\left(Z\right)=7.5X+5Y$

We have to maximize the total profit of manufacturer.

After plotting all the constraints give by equation (1), (2), (3) and (4), we get the feasible region as shown in the image.

Corner points	Value of $Z=7.5X+5Y$
A (0, 40)	200
B (15, 30)	262.50 (Maximum)
C (20, 20)	250
D (20, 0)	150

Hence maximum profit that manufacturer can earn is $262.50Rs$ when 15 toys of type A and 30 toys of type B manufactured.